Python和ctypes：如何正确传递“ pointer-to-pointer”进入DLL？ [英] Python and ctypes: how to correctly pass "pointer-to-pointer" into DLL?

问题描述

我有一个分配内存并返回它的DLL。 DLL中的功能如下：

I have a DLL that allocates memory and returns it. Function in DLL is like this:

void Foo( unsigned char** ppMem, int* pSize )
{
  * pSize = 4;
  * ppMem = malloc( * pSize );
  for( int i = 0; i < * pSize; i ++ ) (* ppMem)[ i ] = i;
}

此外，我有一个python代码，可从我的DLL访问此函数： / p>

Also, i have a python code that access this function from my DLL:

from ctypes import *
Foo = windll.mydll.Foo
Foo.argtypes = [ POINTER( POINTER( c_ubyte ) ), POINTER( c_int ) ]
mem = POINTER( c_ubyte )()
size = c_int( 0 )
Foo( byref( mem ), byref( size ) ]
print size, mem[ 0 ], mem[ 1 ], mem[ 2 ], mem[ 3 ]

我希望 print 将显示 4 0 1 2 3，但它显示 4 221 221 221 221 O_O。任何提示我在做什么

I'm expecting that print will show "4 0 1 2 3" but it shows "4 221 221 221 221" O_O. Any hints what i'm doing wrong?

推荐答案

发布实际代码。C/ C ++代码无法编译为C或C ++。Python代码具有语法错误（使函数调用Foo终止）。下面的代码有效。修复语法和编译器错误后的主要问题是声明函数 __ stdcall ，因此 windll 可以在Python代码中使用。另一个选择是使用 __ cdecl （通常是de错误），并在Python代码中使用 cdll 而不是 windll 。

Post actual code. The C/C++ code doesn't compile as either C or C++. The Python code has syntax errors (] ending function call Foo). The code below works. The main issue after fixing syntax and compiler errors was declaring the function __stdcall so windll could be used in the Python code. The other option is to use __cdecl (normally the default) and use cdll instead of windll in the Python code.

#include <stdlib.h>

__declspec(dllexport) void __stdcall Foo(unsigned char** ppMem, int* pSize)
{
    char i;
    *pSize = 4;
    *ppMem = malloc(*pSize);
    for(i = 0; i < *pSize; i++)
        (*ppMem)[i] = i;
}

demo.py

demo.py

from ctypes import *
Foo = windll.mydll.Foo
Foo.argtypes = [POINTER(POINTER(c_ubyte)),POINTER(c_int)]
mem = POINTER(c_ubyte)()
size = c_int(0)
Foo(byref(mem),byref(size))
print size.value,mem[0],mem[1],mem[2],mem[3]

输出

Output

4 0 1 2 3

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